Small math help with polynomials If one solution of the equation $3x^2 = 8x + 2k + 1$ is $7$ times the other. Find the solutions and the value of $K$.
Note: This isn't a homework question. I'm skipping ahead in my textbook.
Thank you for the help.
Also please help me with this question no.2 also.
Find the other zeroes of $2x^4 - 3x^3 -3x^2 + 6x - 2$ if $-\sqrt{2}$ and $\sqrt{2}$ are two zeroes of the given polynomial
 A: $\bigstar$ First question:
First we form the standard notation:$3x^2-8x-(2k+1)=0$
We know that the sum of the roots of a quadratic equation of the form $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $x_1+x_2 =\frac{8}{3}$
On the other hand: $x_1=7\times x_2$
Solving this two equations two unknowns yields:$x_2 = \frac13$ and $x_1=\frac73$
We also know that the product of the roots of a quadratic equation of the form $ax^2+bx+c$ is  : $x_1 \times x_2=\frac{c}{a}$
so $$x_1 \times x_2 = \frac79 = -\frac{(2k+1)}{3} \implies k=-\frac 53  $$  
$\bigstar$ Second question:
We can write a polynomial as: $P(x)=(x-x_1)(x-x_2)...(x-x_n)$ where $x_1$ to $x_n $ are the zeros of $P(x)$
So given two roots we have:
$$\begin{align}
P(x)&=2x^4 - 3x^3 -3x^2 + 6x - 2\\
&=(x-\sqrt2)(x+\sqrt2)(x-x_1)(x-x_2)\\
&=(x^2-2)(x-x_1)(x-x_2)
\end{align}$$
Dividing P by $x^2-2$ yields:
$$\begin{align}
P(x)&=(x^2-2)(2x^2-3x+1)\\
&=(x^2-2)(x-1)(2x-1)
\end{align}$$
So:
$$x_1=1 , x_2=0.5$$
A: Rearranging the equation, $3x^2-8x-(2k+1)=0$
If $a,7a$ are the solutions, $a+7a=\frac{8}{3}\implies  a=\frac{1}{3}$
So, $a\cdot 7a=-\frac{2K+1}{3}\implies \frac{7}{9}=-\frac{K+1}{3}\implies K=-\frac{5}{3}$
A: Hint: Study this link. Vieta's formulas may prove useful later in your mathematical life. 
A: Hint $\ $ Vieta $\Rightarrow$ roots of $\rm\:x^4 \color{#0A0}{-3/2}\, x^3 - 3/2\, x^2 + 3\,x \color{#C00}{- 1} = 0\:$ have sum $\,\color{#0A0}{3/2},\:$ and  product $\,\color{#C00}{-1}.\:$ 
Therefore $\rm\: \sqrt{2} -\sqrt{2} + r + s = 3/2,\,$ and $\rm\, -\sqrt{2}\,\sqrt{2}\, r\,s = -1,\ $ so $\rm\ r+s =\color{#0A0}{ 3/2},\,$ and $\rm\, r\,s = \color{blue}{1/2}.\:$ Therefore, applying Vieta again,  $\rm\:r,s\:$ are the roots of $\rm\:x^2 - \color{#0A0}{3/2}\, x + \color{blue}{1/2} = 0,\:$ which are $\rm\:1,\ 1/2.$
